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satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1method one solve \[x^3=8\] \[x^38=0\] \[(x2)(x^2+2x+4)=0\] quadratic formula will give the solutions to \[x^2+2x+4=0\]

scottman
 2 years ago
Best ResponseYou've already chosen the best response.0I see and so 2 is one root and the other 2 complex numbers will be the other 2 roots?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1method two solve \(x^3=1\) and multiply by two since one solution is 1, divide the unit circle in two three equal parts one is at 1, the other two are \(\frac{1}{2}+\frac{\sqrt{3}}{2}i\) and its conjugate \(\frac{1}{2}\frac{\sqrt{3}}{2}i\) multiply these numbers by 2

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1to answer your question, yes, the solution to the quadratic will give you the two complex roots

scottman
 2 years ago
Best ResponseYou've already chosen the best response.0ok thanks a lot, you helped me greatly
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